Energy in SHM

The total energy of a particle executing SHM is $E$. If the amplitude is doubled and the frequency is halved, the new total energy will be

There is a body having mass $m$ and performing SHM with amplitude '$a$'. There is a restoring force $F = -Kx$, where $x$ is the displacement. The total energy of the body depends upon which of the following?

For a particle executing S.H.M., the potential energy is $n$ times the kinetic energy when its displacement from mean position is $\left(\frac{2\sqrt{2}}{3}\right)A$, where $A$ is the amplitude of S.H.M. The value of $n$ is

For a particle performing linear S.H.M. of amplitude 'r', the potential energy is '$\lambda$' times its total energy. The displacement of particle is

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of S.H.M., the value of '$x$' is

The maximum potential energy of a block executing S.H.M. is E and amplitude of oscillation is 'A'. The kinetic energy of the block at amplitude A/2 is

For a particle P performing S.H.M., when displacement is 'x', potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between x, E and F is

The amplitude of a particle executing SHM is $3$ cm. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is (in cm)

For a simple pendulum, having time period $T$, the variation of kinetic energy (K.E.) with time $(t)$ is represented by:

A linear harmonic oscillator of force constant $2 \times 10^{6}$ N/m has amplitude $0.01$ m. Its

A mass $0.4$ kg performs S.H.M. with a frequency $\frac{16}{\pi}$ Hz. At a certain displacement it has kinetic energy $2$ J and potential energy $1.2$ J. The amplitude of oscillation is

Maximum kinetic energy of a particle of mass $1$ kg in S.H.M. is $8$ J and the time period is $4$ s. Maximum potential energy during the motion is $10$ J. Then

A particle of mass '$m$' is executing S.H.M. about the origin on x-axis with frequency $\sqrt{\frac{Ka}{\pi m}}$, where K is a constant and a is the amplitude of S.H.M. If '$x$' is the displacement of a particle at time '$t$', the potential energy of a particle will be

A block is fastened to a horizontal spring. The block is pulled to a distance $x = 10$ cm from its equilibrium position (at $x = 0$) on a frictionless surface from rest. The kinetic energy of the block at $x = 5$ cm is $0.25$ J. The spring constant of the spring is nearly (in Nm$-1$)

A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of kinetic energy to potential energy of the particle at time $t = T/12$ is

A particle starts from mean position and performs S.H.M. with period $6$ second. At what time its kinetic energy is $50\%$ of total energy? ($\cos 45^\circ = 1/\sqrt{2}$)

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A\sin\omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$. The value of $\beta$ is ______.

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz. [ take $\pi = \frac{22}{7}$ ]

A particle is executing S.H.M. of amplitude ' $A$ '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m. When the particle passes through the mean position, its kinetic energy is $16 \times 10^{-3}\text{ J}$. The equation of motion of this particle is (Initial phase of oscillation $= 0^\circ$)

An object of mass 0.2 kg executes simple harmonic oscillations along the x -axis with frequency of $\left(\frac{25}{\pi}\right)\text{Hz}$. At the position $x = 0.04\text{ m}$, the object has kinetic energy 1 J and potential energy 0.6 J. The amplitude of oscillation is

For a particle performing S.H.M.; the total energy is ' $n$ ' times the kinetic energy, when the displacement of a particle from mean position is $\frac{\sqrt{3}}{2}A$, where A is the amplitude of S.H.M. The value of ' $n$ ' is

A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $t = \frac{T}{6}$, the ratio of the potential energy to kinetic energy of the particle is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \sqrt{3}/2\right]$

A particle oscillates along the $x$-axis according to the law, $x(t) = x_0\sin^2\left(\frac{t}{2}\right)$ where $x_0 = 1\text{ m}$. The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph

In simple harmonic motion, the total mechanical energy of given system is $E$. If mass of oscillating particle $P$ is doubled then the new energy of the system for same amplitude is:

A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$, where $x =$ _________.

The kinetic energy of a particle, executing simple harmonic motion is 16 J when it is in mean position. If amplitude of motion is 25 cm and the mass of the particle is 5.12 kg, the period of oscillation is

An object of mass $0.2\text{ kg}$ executes simple harmonic motion along $x$ axis with frequency of $\left(\frac{25}{\pi}\right)\text{Hz}$. At the position $x = 0.04\text{ m}$ the object has kinetic energy $0.5\text{ J}$ and potential energy $0.4\text{ J}$. The amplitude of oscillation is ________ $\text{cm}$.

In S.H.M. the displacement of a particle at an instant is $Y = A\cos 30^\circ$, where $A = 40\text{ cm}$ and kinetic energy is 200 J. If force constant is $1 \times 10^{x}\text{ N/m}$, then x will be $\left(\cos 30^\circ = \sqrt{3}/2\right)$

For a body performing simple harmonic motion, its potential energy is $E_x$ at displacement x and $E_y$ at displacement y from mean position. The potential energy $E_0$ at displacement $(x+y)$ is

Starting from mean position, a body oscillates simple harmonically with a period ' $T$ '. After what time will its kinetic energy be $75\%$ of the total energy? $(\sin 30^\circ = 0.5)$

A particle starts oscillating simple harmonically from its equilibrium position with time period ' T '. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? $\left(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\right)$

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at certain displacement '$x$' from the mean position. If '$A$' is the amplitude of S.H.M the value of '$x$' is

The amplitude of a particle executing S.H.M. is $3\text{ cm}$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is

A particle is vibrating in S.H.M. with an amplitude of $4\text{ cm}$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

A particle executes SHM of amplitude A. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

The maximum potential energy of a block executing simple harmonic motion is $25\text{ J}$. A is amplitude of oscillation. At $A/2$, the kinetic energy of the block is

The amplitude of a particle executing SHM is $3\text{ cm}$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: __________ $\text{cm}$

Which graph represents the difference between total energy and potential energy of a particle executing SHM vs its distance from mean position?

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement $(x)$ starting from mean position to extreme position (A) is given by

A body is executing a linear S.H.M. Its potential energies at the displacement '$x$' and '$y$' are '$E_1$' and '$E_2$' respectively. Its potential energy at displacement $(x+y)$ will be

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A\cos(30^\circ)$. If amplitude is $40\text{ cm}$ and kinetic energy at that time is $200\text{ J}$, the value of force constant is $1.0 \times 10^{x}\text{ N m}^{-1}$. The value of $x$ is ____________.

A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is $50\%$ of total energy?

$\left(\cos 45^\circ = \frac{1}{\sqrt{2}}\right)$

The general displacement of a simple harmonic oscillator is $x = A\sin\omega t$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t = \frac{T}{\beta}$. The value of $\beta$ is ______________.

A particle starts oscillating simple harmonically from its mean position with time period '$T$'. At time $t = \frac{T}{12}$, the ratio of the potential energy to kinetic energy of the particle is

$\left(\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \sqrt{3}/2\right)$

Match List-I with List-II

Choose the correct answer from the options given below

A body of mass '$m$' performs linear S.H.M. given by equation $x = P\sin\omega t + Q\sin\left(\omega t + \frac{\pi}{2}\right)$. The total energy of the particle at any instant is

If the amplitude of linear S.H.M. is decreased then

For a body executing S.H.M.:

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal. (3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period. Choose the most appropriate option from the options given below:

A particle performing S.H.M. when displacement is '$x$', the potential energy and restoring force acting on it are denoted by '$E$' and '$F$' respectively. The relation between $x, E$ and $F$ is

A body is performing S.H.M. of amplitude 'A'. The displacement of the body from a point where kinetic energy is maximum to a point where potential energy is maximum, is

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $\frac{3E}{4}$ then its displacement 'y' is given by:

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.

The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

A mass $0.4\text{ kg}$ performs S.H.M. with a frequency $\frac{16}{\pi}\text{ Hz}$. At a certain displacement it has kinetic energy $2\text{ J}$ and potential energy $1.2\text{ J}$. The amplitude of oscillation is

A body is executing S.H.M. under the action of force having maximum magntude $50\text{ N}$. When its energy is half kinetic and half potential; the magnitude of the force acting on the particle is

An object executes SHM along $x$-axis with amplitude $0.06\text{ m}$. At certain distance '$x$' metre from mean position, it has kinetic energy $10\text{ J}$ and potential energy $8\text{ J}$. the distance '$x$' will be

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $t = \frac{T}{4}\text{ s}$ starting from mean position. Assume that the initial phase of the oscillation is zero.

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm${}^{-1}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $\frac{T}{x}$. The value of x is _____________.

For a particle performing SHM when displacement is $x$, the potential energy and restoring force acting on it is denoted by $E$ and $F$, respectively. The relation between $x, E$ and $F$ is

The total energy of a simple harmonic oscillaior is proportional to

The quantity which does not vary periodically for a particle performing SHM is

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be:

A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin $\frac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at t = 210 s will be:

A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy – time graph of the particle will look like:

For a simple pendulum, a graph is plotted between its kinetic energy $(KE)$ and potential energy $(PE)$ against its displacement $d.$ Which one of the following represents these correctly? $\left(\textit{graphs are schematic and not drawn to scale}\right)$

The particle executing simple harmonic motion has a kinetic energy K0cos2$\omega$t. The maximum values of the potential energy and the total energy are respectively

A particle of mass $m$ executes simple harmonic motion with amplitude a and frequency $v.$ The average kinetic energy during its motion from the position of equilibrium to the end is

Starting from the origin a body oscillates simple harmonically with a period of $2\text{ s}$ After what time will its kinetic energy be $75\%$ of the total energy?

The total energy of particle, executing simple harmonic motion is

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is

A body executes simple harmonic motion. The potential energy $(P.E),$ the kinetic energy $(K.E)$ and total energy $(T.E)$ are measured as a function of displacement $x.$ Which of the following statements is true?

A particle of mass m oscillates with simple harmonic motion between points x1 and x2, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph

In a simple harmonic oscillator, at the mean position

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is

The total energy of particle performing SHM depend on

A linear harmonic oscillator of force constant $2 \times 10^{6}\text{ N/m}$ and amplitude $0.01\text{ m}$ has a total mechanical energy of $160\text{ J}$. Its

In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic?

A loaded vertical spring executes S.H.M. with a time period of $4\text{ sec}$. The difference between the kinetic energy and potential energy of this system varies with a period of

A body executes simple harmonic motion with an amplitude $A$. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy?